Question

Gaseous $N_2{O}_4$ dissociates into gaseous ${NO}_2$ according to the reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$. At $300K$ and $1atm$ pressure, the degree of dissociation of $N_2{O}_4$ is $0.2$. If one mole of $N_2{O}_4$ gas is contained in a vessel, then the density of the equilibrium mixture is:

• A. $1.56g/L$
• B. $6.22g/L$
• C. $4.56g/L$
• D. $3.11g/L$

Answer 1

Answer: (D)

$3.11g/L$

Molecular weight of the mix$=x_{N_2{O}_4}{M}_{N_2{O}_4}+x_{N{O}_2}{M}_{N{O}_2}$---------1

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

At equilibrium, $1-0.2\rightleftharpoons 2\times 0.2$

$x_{N{O}_2}=\frac{0.4}{0.4+0.8}=\frac{1}{3}$

$x_{N_2{O}_4}=\frac{0.8}{0.4+0.8}=\frac{2}{3}$

$\therefore \left(M{W}_{mix}\right)=\frac{2}{3}\times 92+\frac{1}{3}\times 46=76.67$

$\because PV=\frac{m}{M}RT\Rightarrow PM=dRTord=\frac{PM}{RT}$

$\therefore d=\frac{1\times 76.67}{0.0821\times 300}=3.11g/L$